# Rectangle hyperbola

#### arze

The tangent at the point on the curve $$\displaystyle x=ct$$ , $$\displaystyle y=\frac{c}{t}$$ meets the X and y axes at P and Q respectively. The normal at the point meets the line y=x and y=-x at R and S respectively. Prove that PRQS is a rhombus unless $$\displaystyle t^2=1$$

I found
$$\displaystyle P(2ct,0)$$ , $$\displaystyle Q(0,\frac{2c}{t})$$
and $$\displaystyle R(\frac{c(t^2+1)}{t},\frac{c(t^2+1)}{t})$$
$$\displaystyle S(\frac{c(t^2-1)}{t},-\frac{c(t^2-1)}{t})$$

I tried the gradients of the diagonals PQ and RS are perpendicular, and got no restrictions on $$\displaystyle t^2$$ I also tried finding the lengths of all the sides but they are all the same $$\displaystyle \frac{c}{t}\sqrt{t^4+1}$$ so i cannot prove that PRQS is a rhombus unless $$\displaystyle t^2=1$$.
Thanks

#### sa-ri-ga-ma

At t^2 = 1, S co-ordinates are (0, 0).

Normal at any point on the hyperbola will not pass through the origin, unless it is drawn at the vertex. In that case, R and S will coincide and tangent at P will not cut y-axis.

• arze

#### arze

is there any like mathematical proof of this as in like from equations and such?
thanks